# Finite group with a character having one nonzero absolute value

Let $G$ be a finite group. Assume that $\chi$ is a complex irreducible character of $G$ of degree $n\geq 2$, with the property that for each element $g\in G$ either $\chi(g)=0$ or $|\chi(g)|=n$.

1. Does it necessarily follow that $\chi$ is imprimitive, i.e., induced from a character of a subgroup?

The only such examples I could think of are given by extraspecial groups.

1. Is there a classification of all the finite groups with this property?

The questions are partly motivated by this post: Finite groups with a character having very few nonzero values?

• There is a whole literature on finite groups of central type, which is the name for the characters you describe. – Geoff Robinson Aug 8 '18 at 11:44
• I believe that any finite group with such a faithful character is solvable, and I believe the result is due to Howlett and Isaacs. – Geoff Robinson Aug 8 '18 at 11:54
• Thanks! I’m sure it’s probably obvious, but is there a quick way to see that my condition is equivalent to saying that $G$ has a character of degree equal to the square root of the index of the center? This seems to be the usual way groups of central type are defined. – Goro Aug 8 '18 at 12:10
• Just compute the character inner product $\langle \chi, \chi \rangle = 1$, which means that $\sum_{g \in G} |\chi(g)|^{2} = |G|$. On the other hand,$|\chi(g)|^{2} = \chi(1)^{2}$ for $g \in Z(G),$ and $0$ otherwise, so we obtain $|Z(G)|\chi(1)^{2} = |G|$ and $\chi(1)^{2} = [G:Z(G)].$ On the other hand, if $G$ has a faithful irreducible character $\mu$ with $\mu(1)^{2} = [G:Z(G)],$ then $\mu$ must vanish identically outside $Z(G)$ ( and we have $|\mu(z)| = \mu(1)$ for $z \in Z(G)).$ – Geoff Robinson Aug 8 '18 at 13:28
• I won't write this as a (partial) answer, but here are a couple of general properties of groups of central type: If G is of central type, then so is every Sylow p-subgroup ( a Sylow p-subgroup $P$ might be Abelian, in which case $P \leq Z(G)$, and $P$ is a direct factor of $G$. On the other hand, if every Sylow p-subgroup of $P$ $G$ is of central type (for every choice of $p$) with $Z(P) \leq Z(G),$ then $G$ is of central type ( this needs Brauer'c characterization of characters). Also the faithful character $\chi$ is in a $p$-block of full defect for every $p$. – Geoff Robinson Aug 9 '18 at 10:49

Using this result, we can show that a character $\chi$ of central type is necessarily imprimitive. For suppose that $Z$ is the center of $\chi$, so that $\chi(1)^2 = \lvert G:Z \rvert$, and let $N$ be any normal subgroup of $G$ containing $Z$. Let $\tau$ be a constituent of $\chi_N$. By Clifford theory, $\chi$ is induced from the inertia subgroup of $\tau$ in $G$. So if $\tau$ is not invariant in $G$, we are done. Otherwise, we have $\chi_N = e\tau$ and $\tau$ vanishes outside $Z$. It follows that $e^2 = \lvert G:N \rvert$ and $\lvert G:N \rvert$ is a square. But by solvability of $G$, maximal normal subgroups have prime index.
To the best of my knowledge, groups of central type are not classified. The paper by Howlett and Isaacs contains a non-nilpotent example. There are also nilpotent examples of arbitrarily large nilpotency class: Take the semidirect product of a cyclic group of order $p^{n+1}$ with the Sylow $p$-subgroup of its automorphism group.